正则表达式日期格式小于运算符

Question

I am trying to use regex to verify a date format and I would like to check if the day is less than 32. Similarly, that the month is also less than 12. I have no idea how to about it. Currently, this is what I have;

^[0-1]?[0-9]{1}\-[0-3]?[0-9]{1}\-[0-9]{2,4}$

This regex achieves the format (m)m-(d)d-(yy)yy

Answer 1

TL;DR

Don't use regular expressions for comparison operations. Use a regex to split off values to compare, or use an actual parser.

Use Regular Expressions to Extract Comparables

Date comparisons is a really poor problem for regex to solve. At most, you should use a regular expression to extract your days of the month for a numeric comparison. For example:

date = '01-01-1970'
date.split('-')[1].to_i < 32
#=> true

However, the code above won't really tell you if a given date is valid . For example, what about February 30th or November 31st? Instead, you should attempt to parse the date to determine its validity.

Use a Date Parser

The best way to tell if a given date is valid is to parse it with a date parser, and then report a Boolean result or handle the exception. For example, you could attempt to parse the date with Date#parse .

Boolean Results

If you just want a Boolean result, you can coerce a valid/invalid parse to true or false. For example:

require 'date'

date = '01-33-1970'
!!(Date.parse date rescue nil)
#=> false

Rescuing and Reporting the Exception

Less magically, you would need to rescue ArgumentError from Date#parse . For example:

require 'date'

def valid_date? date_string
  true if Date.parse date_string
rescue ArgumentError => e
  STDERR.puts "#{e.class}: #{e}: '#{date_string}'"
  false
end

valid_date? '11-31-1970'

This will do what you expect, albeit more verbosely. For example, the above example will print the exception to standard error, and then return false as the result.

ArgumentError: invalid date: '11-31-1970'
#=> false

Answer 2

^(?:[0-1][1-2]|[1-9])\-(?:3[0-1]|[0-2][1-9]|[1-9])\-[0-9]{2}(?:[0-9]{2})?$

should do what you're looking for. It will only allow months from 1-12 (either 1-9 or 01-12), days from 1-31 (either 1-9 or 01-31) and years of at least 2 digits with a maximum of four. Tested on regex101.

Answer 3

Basic:

Here is a regex that should do what you want:

^(0[1-9]|1[0-2]|[1-9])-(0[1-9]|[1-2][0-9]|3[0-1]|[1-9])-\d{2}(\d{2})?$

It matches months greater than 0 and less than 13, then - , then days greater than 0 and less than 32, then - , then years (2 digits or 4 digits).

Bonus:

Full regex for matching dates in that format with validation:

^((0?[13578]|10|12)-(([1-9])|(0[1-9])|([12])([0-9]?)|(3[01]?))-((19)([2-9])(\d{1})|(20)([01])(\d{1})|([8901])(\d{1}))|(0?[2469]|11)-(([1-9])|(0[1-9])|([12])([0-9]?)|(3[0]?))-((19)([2-9])(\d{1})|(20)([01])(\d{1})|([8901])(\d{1})))$

Answer 4

If you want to determine the string is a valid date, you'd be better off attempting to convert it. If it won't convert, it's not valid.

def date_valid?(date_string)
  format = '%m/%d/' + (date_string.split(-).last.size == 4 ? '%Y' : '%y')
  return true if Date.strptime(date_string, format)
  rescue ArgumentError
  return false
end